1950 AHSME Problems/Problem 48

Problem

A point is selected at random inside an equilateral triangle. From this point perpendiculars are dropped to each side. The sum of these perpendiculars is:

$\textbf{(A)}\ \text{Least when the point is the center of gravity of the triangle}\qquad\\ \textbf{(B)}\ \text{Greater than the altitude of the triangle} \qquad\\ \textbf{(C)}\ \text{Equal to the altitude of the triangle}\qquad\\ \textbf{(D)}\ \text{One-half the sum of the sides of the triangle} \qquad\\ \textbf{(E)}\ \text{Greatest when the point is the center of gravity}$

Solution

begin by drawing the triangle, the point, the altitudes from the point to the sides, and the segments connecting the point to the vertices. Let the triangle be $ABC$ with $AB=BC=AC=s$ . We will call the aforementioned point $P$ . Call altitude from $P$ to $BC$ $PA'$ . Similarly, we will name the other two altitudes $PB'$ and $PC'$ . We can see that $\frac{1}{2}sPA'+\frac{1}{2}sPB'+\frac{1}{2}sPC'=\frac{1}{2}sh$ Where h is the altitude. Multiplying both sides by $2$ and dividing both sides by $s$ gives us $PA'+PB'+PC'=h$ The answer is $\textbf{(C)}$

1950 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 47	Followed by Problem 49
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1950 AHSME Problems/Problem 48

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